Question

I just learned how to derive the quadratic formula from ax2 bx+c, but I can't figure out how to derive -b/2a (the x coordiante for a vertex). Any ideas?

Answer 1

complete the square

Answer 2

that comes from completing the square and writing
\[y=ax^2+bx+c\] as
\[y=a(x+\frac{b}{2a})^2+k\] and since the first term is a perfect square, if "a" is positive the least it can be is 0, and if a is negative the greatest it can be is 0, leaving a max of min of y at k. this occurs when
\[x=-\frac{b}{2a}\]

Answer 3

I'm confused... where did c go? and where did k come from?

Answer 4

\[ax^2+bx+c=0\]
\[x^2+\frac{b}{a}x+\frac{c}{a}=0\]completing th esquare
\[x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=-\frac{c}{a}+\frac{b^2}{4a^2}\]
\[x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=\frac{b^2-4ac}{4a^2}\]
\[(x+\frac{b}{a})^2=\frac{b^2-4ac}{4a^2}\]
\[(x+\frac{b}{a})^2=\pm\frac{\sqrt{b^2-4ac}}{2a}\]
subtract b/a from both sides
\[x=-\frac{b}{a}\pm\frac{\sqrt{b^2-4ac}}{2a}\]
simplifying to get

\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Answer 5

Whoah, thanks for that hard work! My brain is currently processing...

Answer 6

No, wait, I already know how to get the quadratic formula, I want to know how to get
-b/2a from it

Answer 7

If you let let the discriminant \[b^2-4ac=0\] then
\[b^2=4ac\] and
\[x=\frac{-b\pm\sqrt{b^2-b^2}}{2a}\rightarrow\frac{-b}{2a}\]

Answer 8

Read again what sat73 wrote carefully. He wrote k for all the stuff left over, because it's exact value isn't important. For the record, \(k = (b^2-4ac)/2\) but as I say that's not critical for the argument.

Having showing that the equation equals
\[a(x +b/2a)^2 + k\]
you now ask yourself when does it have a max or min; i.e., where is the vertex.

If occurs precisely when the first term is zero, i.e.,
\[a(x +b/2a)^2 = 0\]
which means \(x = -b/2a\)

Answer 9

You could also take the first derivative and set it equal to zero to find the critical point
\[\frac{d}{dx}=2ax+b=0\]
\[x = -\frac{b}{2a}\]

Answer 10

careful cyter you are using calculus to explain it and the student might not be in calculus at all (perhaps Algebra I or II)